How do I integrate a polynomial with a variable in the denominator?

In summary: You could replace it with Ex, but then you'd have to use another symbol, say F. Then you have a choice: 1) you could use E+F to replace Cx+D, or 2) you could use C to replace Cx+D. Both methods are equivalent. I hope that explains it!In summary, the conversation discusses the correct way to integrate the polynomial \frac{dy}{dx} = \frac{3y^2}{x}. The correct integral is \frac{1}{3}\int\frac{dy}{y^2} = \int x^{-1}dx, which simplifies to \frac{1}{3}\ln|y^2| = \ln |
  • #1
snowJT
117
0
I Just want to know if this is how this should be done..

[tex]\frac{dy}{dx} = \frac{3y^2}{x}[/tex]

[tex]dy = \frac{3y^2}{x}dx[/tex]

[tex]dy = 3^2 x^-^1dx[/tex]

[tex]\frac {1}{3} \int \frac {dy}{y^2} = \int x^-^1dx[/tex]

[tex]\frac {1}{3} ln| y^2| = ln | x | + C[/tex]

because I forget how to integrate the polynomial...
 
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  • #2
The right hand side is correct. Look again at [tex]\int y^{-2}dy[/tex] How do you integrate this?
 
  • #3
[tex]\frac {1}{3} \int \frac {dy}{y^2} = \int x^-^1dx[/tex]

That's OK, but the integral of 1/y^2 isn't ln(y^2).

Look up how to integrate polynomials, if you forgot. Or if you know how to differentiate polynominals, work it out from the fact that integration is "anti-differentiation".
 
  • #4
its not [tex]\frac{1}{3y}[/tex] is it?
 
  • #5
snowJT said:
its not [tex]\frac{1}{3y}[/tex] is it?

Close; it's -1/(3y), since you must divide by the power of y(which is -1)
 
  • #6
oops.. thanks I was thinking that was how you did it originally, I was just confused and did it that other way and forgot ket things... thanks

EDIT: Hmmm I also need to then solve for Y

[tex]y^-^1 = \frac {-ln|x|}{3} - \frac {C}{3}[/tex]

[tex]y = \frac{-3}{ln|x|} - \frac{3}{C}[/tex]
 
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  • #7
that looks wrong to me
 
  • #8
not even close i don't think
 
  • #9
[tex]y=\frac{ln x + C}^-1{3}[/tex]
 
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  • #10
snowJT said:
EDIT: Hmmm I also need to then solve for Y

[tex]y^{-1} = \frac {-ln|x|}{3} - \frac{C}{3}=\frac{-(ln|x|+C)}{3}[/tex]

[tex]y=\frac{-3}{(ln|x|+C)}[/tex]

Can you follow what I've done?

JJ420 said:
that looks wrong to me

JJ420 said:
not even close i don't think

JJ420; we are here to help, not criticise!
 
  • #11
[tex]y=\frac{ln x + C}^-1{3}[/tex]
thats what i got but I am dumb so don't trust me
 
  • #12
oops that -13 is supposed to be an exponent of the numerator (-1) and 3 in the denominator
 
  • #13
JJ420 said:
[tex]y=\frac{(ln x + C)^{-1}}{3}[/tex]
thats what i got but I am dumb so don't trust me

[tex]\frac{(ln x + C)^{-1}}{3}=\frac{1}{3(ln x + C)}[/tex] The 3 should definitely be in the numerator, and we need a - sign too. See my post:

cristo said:
[tex]y=\frac{-3}{(ln|x|+C)}[/tex]
 
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  • #14
cristo said:
[tex]y^{-1} = \frac {-ln|x|}{3} - \frac{C}{3}=\frac{-(ln|x|+C)}{3}[/tex]

[tex]y=\frac{-3}{(ln|x|+C)}[/tex]

Can you follow what I've done?

no I can't figure it out, common denominator?
 
  • #15
oh.. yes I see

why is C positive and not negative
 
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  • #16
snowJT said:
no I can't figure it out, common denominator?

Yea you just write ln|x| and C as one fraction with denominator 3. Then take the reciprocal of both sides (since the LHS is 1/y) to obtain an expression in terms of y.

snowJT said:
oh! you simplified it before you swapped the fractions! is that a rule?
Well, yes. In order to take the reciprocal of both sides, there must only be one fraction on each side.
 
  • #17
looking back to the origonal i don't see how there is a negative
 
  • #18
JJ420 said:
looking back to the origonal i don't see how there is a negative

The minus sign comes about when integrating; integral(y-2)=-y-1 (+C)
 
  • #19
snowJT what does C mean?

it means any constant...it can be a minus but since we don't know we just write + C
 
  • #20
sara_87 said:
snowJT what does C mean?

it means any constant...it can be a minus but since we don't know we just write + C

figured so, there were some examples I just found in my book
 
  • #21
snowJT said:
why is C positive and not negative

Because you defined it as positive! When you integrated you wrote +c on the RHS, then multiplied through with the minus sign.
 
  • #22
snowJT said:
figured so, there were some examples I just found in my book

you could find the answer in your books if you just read with an open mind, but it feels more real, less boring and it is less likely that you'd forget the anwswer when you ask some one else
 
  • #23
cristo said:
Because you defined it as positive!

:rofl: you don't have to shout
 
  • #24
sara_87 said:
:rofl: you don't have to shout

Haha, I didn't. You'd know when I WAS SHOUTING! :tongue2:
 
  • #25
I see you've got that homework helper recognition, is that new?
 
  • #26
sara_87 said:
I see you've got that homework helper recognition, is that new?

Yea, it just popped up yesterday.
 
  • #27
congratulations! I'm jelous but unlike yourself i don't deserve one
(and i won't post up anything unrelated to this thread anymore, appologies snowJT)
 
  • #28
Thanks. Yea, this thread has got very off topic. Sorry, although I think the question is answered now!
 
  • #29
C is always positive and never has a coefficent is the final result because any multiple of some constant is just another constant, and A negative value of C is allowed, we write it Positive because its easier to work with.

Edit: I always do this..didnt see this page, just the first one..omg kill me
 
  • #30
It doesn't matter what sign you use for C, or what letter. C is just an undetermined constant. The standard practice whenever you have undetermined constants is to write them in as simple form as possible. Thus you would choose to write Cx, not -Cx if you had the chance. You can also add them together: C+D can be rewritten as a single new constant E. Also, to save using an unnecessary number of letters being used, we often replace C+D with just another C. This is equivalent to replacing C+D with E, and then, now we've freed up the C we can reuse that symbol by using C instead of E. Note that you cannot simplify Cx+D.
 
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